41. A sailor strikes the side of his ship just below the
surface of the sea. He hears the echo of the wave reflected from the ocean
floor directly below 3.0 s later. How deep is the ocean at this point?

42. S and P waves from an earthquake travel at
different speeds, and this difference helps in the determination of the
earthquake "epicenter" (where the disturbance took place).
(a) Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves,
respectively, how far away did the earthquake occur if a particular seismic
station detects the arrival of these two typed of waves 2.0 min apart? (b)
Is one seismic station sufficient to determine the position of the
epicenter? Explain.

(a) Because two waves travel in the same distance,
therefore, we will have:

(b) Since the waves spread out in a circle, and we are not
certain which direction the waves are traveling, we could not determine the
exact location of the quake. We would need another station not in line
with the other two. (Table of contents)

51. If a violin string vibrates at 440 Hz as its
fundamental frequency, what are the frequencies of the first four harmonics?

53. A particular string resonates in four loops at a
frequency of 280 Hz. Name at least three other frequencies at which it
will resonate.

Four loops means four anti-nodes, which means that we have the fourth
harmonic, as the fundamental is one "loop" or anti-node. The
fourth harmonic would be four times the fundamental, so therefore the
fundamental is 280 Hz/4 = 70 Hz, and the next two would be 140 Hz, and 210
Hz. So
f1 = 70Hz (fundamental frequency)
f2 = 2f1 = 2 x 70 = 140 Hz
f3 = 3f1 = 3 x 70 = 210 Hz
f4 = 4f1 = 4 x 70 = 280 Hz
(Table of contents)

55. The velocity of waves on a string is 92 m/s. If
the frequency of standing waves is 475 Hz, how far apart are two adjacent
nodes?

56. If two successive overtones of a vibrating string are
280 Hz and 350 Hz, what is the frequency of the fundamental?

If it is a vibrating string, we know that the harmonics are successive
multiples of the fundamental, (f1, 2f1, 3f1,
4f1, ...), and so the difference between successive harmonics or
overtones is simply the fundamental. In this case, 350 Hz - 280 Hz = 70
Hz. (Note that these two are the fourth and fifth harmonics
respectively)
(Table of contents)

57. A guitar string is 90 cm long, and has a mass of 3.6
g. From the bridge to the support post (=L) is 60 cm, and the string is
under a tension of 520 N. What are the frequencies of the fundamental and
the first two overtones?

Now we will apply a special formula from the data packet, which is:
v =ÖT/µWhere µ = mass/length for the cord, and T is the tension in the cord.
have: T = 520N; m = 3.6 g = .0036 kg; and l = .90 m, so µ = (.0036 kg)/(.90 m) =
.004 kg/m
Therefore, v = ÖT/µ
= Ö(520
N/(.004 kg/m)) = 360.56 m/sAt the fundamental, the string has two nodes at the end, and one
anti-node in the middle, so therefore the vibrating length of the string (.60
m) is equal to 2/4l, so l
= 1.2 m Now we can get the fundamental by using
v = ¦l f = v/l = (360.56 m/s)/(1.2 m) = 300.5 Hz, and
the next two harmonics are simply multiples of this fundamental:
f1 = 300.5 Hz (fundamental frequency)
f2 = 2f1 = 2 x 300.5 Hz = 601 Hz
f3 = 3f1 = 3 x 300.5 Hz = 901 Hz(Table of contents)

Chapter 12

1. A hiker determined the length of a lake by listening
for the echo of her shout reflected by a cliff at the far end of the lake.
She hears the echo 1.5 s after shouting. Estimate the length of the
lake.

You hear the echo when the sound, traveling at 343 m/s, goes
twice the length of the lake. (i.e. there and back again). It the
sound travels for 1.5 s total, it went (343 m/s)(1.5 s) = 514.5 m total, and
therefore the lake is half that in length, or 257.25 m long (about 260 m)(Table of contents)

3. a) Calculate the wavelengths in air at 20oC
for sounds in the maximum range of human hearing, 20 Hz to 20,000 Hz. b) What is
the wavelength of a 10-MHz ultrasonic wave?

5. A person sees a heavy stone strike the concrete
pavement. A moment later two sounds are heard from the impact: one
travels in the air, and the other in the concrete, and they are 1.4 s
apart. How far away did the impact occur?

The speed of sound in air we will take to be 343 m/s, and
that in concrete is 2950 m/s. (They figure it out using a chapter we
skipped)
In general, the time for the sound to travel is t = s/v, so we have that we
hear the concrete transmitted sound 1.4 seconds before the air transmitted
sound or,
s/va - s/vc = 1.4 s where va and vc
are the speeds of sound in air and concrete. If we solve for s, we get:
s = (1.4 s)/(1/va - 1/vc) = 540 m(Table of contents)

26. The A string on a violin has a fundamental frequency
of 440Hz. The length of the vibrating portion is 32cm and has a mass of 0.35 g .
Under what tension must the string be placed?

The fundamental frequency is: f1 = 440Hz
At the fundamental, the string has two nodes at the end, and one anti-node in
the middle, so therefore the vibrating length of the string (.32 m) is equal
to 2/4l, so l
= .64 m
v = lf = (0.64 m)(440 Hz) = 281.6 m/s

Now, at 440 Hz, the wavelength must bel = v/f = (462 m/s)/(440 Hz) = 1.05 m, and since at
the fundamental, only half the wavelength fits on the string, then we get that
the string must now be (1.05 m)/2 = .525 m long, which means the string must
now be fingered .70 - .525 m = .175 m from the end.
(Table of contents)

28. Determine the length of an open organ pipe that
emits middle C (262Hz) when the temperature is 210 C.

The speed of sound depending on temperature derives from the formula: v =
331 + 0.6T m/s
Thus, the speed of sound now is: v = 331 + 0.6*21 = 343.6 m/s
The wavelength of a 262 Hz sound at this temperature is: l = v/f =
343.6/262 = 1.31m
Since a both ends open organ pipe has a node in the middle, and two anti-nodes
at each end, the length of the pipe (L) is equal to 2/4l,
or L = l/2 = (1.31 m)/2
= 0.66m
(Table of contents)

29. (a) What resonant frequency would you expect from
bowling across the top of an empty soda bottle that is 15 cm deep? (b) How would
that change if it was one-third full of soda?

(a) The empty soda bottle is a closed pipe with the node is at the bottom
of the bottle (fixed end) and the anti-node is at the top. Therefore the
length of the bottle (L) is equal to only 1/4l,
or l=
4L = 4(.15 m) = 0.60 m
The resonant frequency must be: f = v/l =
(343 m/s)/(0.60 m) = 571.67Hz (Assuming the speed of sound is 343 m/s)

(b) The length of the pipe is now 2/3 of what
is was or (2/3)(.015 m) = .40 m
The resonant frequency now is: f = v/l= (343
m/s)/(0.40 m) = 857.5Hz
(Table of contents)

30. If you were to build a pipe organ with open tube
pipes spanning the range of human hearing (20Hz to 20kHz), what would be the
range of lengths of pipes required?

The speed of sound in air is 343m/s.
With f1 = 20 Hz, the wavelength is: l = v/f
= (343 m/s)/(20 Hz) = 17.15 m. Since a both ends open organ pipe has a
node in the middle, and two anti-nodes at each end, the length of the pipe (L)
is equal to 2/4l, or
L = l/2
= (17.15 m)/2 = 8.575m (longest)
With f1 = 20,000 Hz, the wavelength is: l = v/f
= (343 m/s)/(20,000 Hz) = .01715 m. Since a both ends open organ pipe
has a node in the middle, and two anti-nodes at each end, the length of the
pipe (L) is equal to 2/4l,
or L = l/2
= (.01715 m)/2 = .008575 m (smallest)
Therefore, the range of the length must
be:
0.008575m =< L =< 8.575m
(Table of contents)

31. An organ pipe is 112 cm long. What are the
fundamental and first three audible overtones if the pipe is (a) closed at one
end, and (b) open at both ends?

If one end is closed, and one open, then you get a pattern
of successive harmonics that goes
f1, 3f1, 5f1, 7f1, 9f1
- that is, odd multiples of the fundamental, so let's find the fundamental:With one end fixed, or a closed pipe, at the fundamental, the
node is at the closed end (fixed end)
and the anti-node is at the open end. Therefore the length of the pipe (L) is
equal to only 1/4l, or l=
4L = 4(1.12 m) = 4.48 m, and the frequency you can get with v = fl.
The resonant frequency now is: f = v/l= (343 m/s)/(4.48 m) = 75.6
HzAnd the next two can be found now:
f1 = 75.6 Hz (fundamental frequency)
f2 = 3f1 = 3 x 75.6 Hz = 230 Hz
f3 = 5f1 = 5 x 75.6 Hz = 383 Hz
f3 = 7f1 = 7 x 75.6 Hz = 536 Hz

If both ends are open, then you get a pattern of successive
harmonics that goes
f1, 2f1, 3f1, 4f1, 5f1
- that is, multiples of the fundamental, so let's find the fundamental:At the
fundamental, a both ends open organ pipe has a node in the middle, and
two anti-nodes at each end, the length of the pipe (L) is equal to 2/4l,
or L = l/2,
or l= 2L = 2(1.12 m) = 2.24 m, and
the frequency you can get with
v = fl.
The resonant frequency now is: f = v/l= (343 m/s)/(2.28 m) = 153.1
HzAnd the next two can be found now:
f1 = 153.1 Hz (fundamental frequency)
f2 = 2f1 = 2 x 153.1 Hz = 306 Hz
f3 = 3f1 = 3 x 153.1 Hz = 459 Hz
f3 = 4f1 = 4 x 153.1 Hz = 612 Hz(Table of contents)

33. A highway overpass was observed to resonate as one
full loop when a small earthquake shook the ground vertically at 4.0 Hz. The
highway department put a support at the center of the overpass, anchoring it to
the ground as shown in Fig. 12 - 34. What resonant frequency would you now
expect for the overpass? It is noted that earthquake rarely do significant
shaking above 5 or 6 Hz. Did the modifications do any good?

With one support in the center, the second harmonic, or
first overtone could still vibrate or resonate, as it has a node here.
(So could the fourth, the 8th, all octaves (doublings) of the
fundamental). The second harmonic would have twice the frequency, or
vibrate at 8.0 Hz, and since apparently earthquake waves don't go this high in
frequency, it would seem that this would work.(Table of contents)

35. (a) At T = 150C, how long must a close
organ pipe be if it is to have a fundamental frequency of 294 Hz? (b) If this
pipe were filled with helium, what would its fundamental frequency be?

Well, since they give a temperature, we must calculate the
new speed of sound:
v = 331 m/s + (.60 m/s/oC)(15 oC) = 340 m/s.

The wavelength of a 294 Hz wave at this wave speed is l
= v/f = (340 m/s)/(294 Hz) = 1.1565 mWith one end fixed, or a closed pipe, at the fundamental, the
node is at the closed end (fixed end)
and the anti-node is at the open end. Therefore the length of the pipe (L) is
equal to only 1/4l, so L = 1/4(1.1565
m) = .289 m

The speed of sound in Helium is 1005 m/s (Look it up in your book -
table 12-1 p 348) The length of the pipe (L) is equal to only 1/4l,
or l=
4L = 4(.289 m) = 1.1565 m, (we have figured this out
before) and the frequency you can get with
v = fl.
The resonant frequency now is: f = v/l= (1005 m/s)/(1.1565
m) = 869 Hz
(Table of contents)

36. A particular organ pipe can resonate at 264Hz,
440Hz, and 616Hz. (a) Is this an open or closed pipe? (b) What is the
fundamental frequency of this pipe?

A both ends open organ pipe would resonate at multiples of a
fundamental, or f1, 2f1, 3f1, 4f1, 5f1
- and if it had one end closed, it would resonate at
f1, 3f1, 5f1, 7f1, 9f1
- that is, odd multiples of the fundamental. So which is this??

So I think they want us to say that it is a closed end pipe, and we are
seeing the second (3x) third (5x) and fourth (7x) harmonics, but since they
never said that these were successive harmonics (i.e. that there were no other
resonant frequencies in between) then the pipe really could be a both ends
open with a fundamental of say 44 Hz or 22 Hz or 11 Hz or even 2 H or 1 Hz
(That is any thing that would go into (264, 440, 616) evenly - any factor of
88)(Table of contents)

37. A uniform narrow tube 1.8m long is open at both
ends. It resonates at two successive harmonics of frequency 275Hz and 330Hz.
What is the speed of sound in the gas in the tube?

A both ends open organ pipe would resonate at multiples of a
fundamental, or
f1, 2f1, 3f1, 4f1, 5f1
- so successive harmonics are separated in frequency by the fundamental
frequency, 330 Hz - 275 Hz = 55 Hz in this case. Since a both ends open organ pipe has a node in the middle, and
two anti-nodes at each end, the length of the pipe (L) is equal to 2/4l,
or L = l/2,
or l= 2L = 2(1.8 m) = 3.60 m, and
the velocity you can get with
v = fl.
v = (55 Hz)(3.60 m) = 198 m/s(Table of contents)

38. A pipe in air at 200C is to be designed
to produce two successive harmonics at 240Hz and 280Hz. How long must the pipe
be, and is it open or closed?

A both ends open organ pipe would resonate at multiples of a
fundamental, or
f1, 2f1, 3f1, 4f1, 5f1
- and if it had one end closed, it would resonate at
f1, 3f1, 5f1, 7f1, 9f1
- that is, odd multiples of the fundamental. So which is this??

Let's play with numbers:
280 - 240 = 40 Hz

280/40 = 7
240/40 = 6
Since these are successive harmonics (none in between) then this must be a
both ends open pipe with a fundamental frequency of 40 Hz.

The velocity of sound at room temperature is 343 m/s, and at 40 Hz the
wavelength is
= (343 m/s)/(40 Hz) = 8.575 mAt the
fundamental, a both ends open organ pipe has a node in the middle, and
two anti-nodes at each end, the length of the pipe (L) is equal to 2/4l,
or L = l/2,
L = (8.575 m)/2 = 4.3 m(Table of contents)

42. A piano turner hears one beat every 2.0s when
trying to adjust two strings, one of which is sounding 440Hz, so that they sound
the same tone. How far off in frequency is the other string?

If the beat period is 2.0 s, then the beat frequency is .50
Hz (f = 1/T)
We know that the beat frequency is just the difference in the two
frequencies. So the other frequency is either 439.5 or 440 .5
Hz. (Table of contents)

43. What will be the "beat frequency" if
middle C (262Hz) and C# (277Hz) are played together? Will this be audible? What
if each is played two octaves lower (each frequency reduced by a factor of 4)?

The beat frequency would just be the difference in the
frequencies:¦beat
= |¦1
- ¦2|¦beat
= |262 Hz - 277 Hz| = 15 Hz
And no, you would not be able hear that as beats, as the human ear can hear
about 8 beats/sec
Two octaves lower, the frequencies would be:
262/4 = 65.5 Hz
and
277/4 = 69.25 Hz
Again, he beat frequency would just be the difference in the frequencies:¦beat
= |¦1
- ¦2|¦beat
= |69.25 Hz- 65.5 Hz| = 3.75 Hz
And this would be audible as beats.(Table of contents)

44. A certain dog whistle operates at 23.5hHz, while
another (brand X) operates at an unknown frequency. If neither whistle can be
heard by humans when played separately, but a thrill whine of frequency 5000Hz
occurs when they are played simultaneously, estimate the operating frequency of
brand X.

The mystery whistle has a frequency that is 5000 or 5 kHz
from the other, which means is is at 23.5 + or - 5 kHz, which means the two
options are 28.5 kHz (above human hearing) or 18.5 kHz which would be audible,
and since it is not audible, then the other must be the higher frequency of
28.5 kHz.(Table of contents)

45. A guitar string produces 4 beats/s when sounded
with a 350Hz tuning fork and 9 beats when sounded with a 355Hz tuning fork. What
is the vibrational frequency of the string? Explain your reasoning

At 350 Hz, and 4 beats/s, the other string could be 354 or
346 Hz, and at 355 Hz, and 9 beats/s, the other string could be 346 Hz, or 364
Hz. The only frequency that would work or course is 346 Hz.(Table of contents)

48. Two loudspeakers are 2.5 m apart. A person
stands 3.0 m from one speaker, and 3.5 m from the other. a) What is the
lowest frequency at which destructive interference will occur at this
point? b) calculate two other frequencies that also result in destructive
interference at this point. (give the next two highest)

If destructive interference occurs at this point, then the
difference in distance from one speaker to the next (.50 m in this case) must
be have a half wavelength remainder. (i.e. the distance is 1/2l,
1 1/2l,
2 1/2l,
3 1/2l ...)
The lowest frequency would be the longest wavelength, so
the difference in dis
.50 m = 1/2l
l
= 1.0 m
Assuming that v for sound is 343 m/s, we can usev = ¦lso f = 343 Hz

The next would be the next longest
.50 m =1 1/2l
l
= .3333 m
Assuming again that v for sound is 343 m/s, we can usev = ¦lso f = 1029 Hz

The last would be the next longest
.50 m =2 1/2l
l
= .2 mv = ¦lso f = 1715 Hz(Table of contents)

49. A source emits sound of wavelengths 2.64 m and 2.76
m
in air. (a) How many beats per second will be heard (assume T = 200C)?
(b) How far apart in space are the regions of maximum intensity?

21. For the wave y = 5sin30p[t-(x/240)],
where x and y are in centimeters, and t is in seconds, find the a) displacement
when t = 0, and x = 2 cm; b) wavelength; c) velocity of the wave, and d)
frequency of the wave

23. For the wave shown in Fig 23-3, find its
amplitude, frequency, and wavelength if its speed is 300 m/s. Write the
equation for this wave as it travels out along the +x axis if its position at t
= 0 is as shown